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On these pages, we present most recent results of our joint research, directly from the lab. |
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Basic bibliography:
K. Belabas, A fast algorithm to compute cubic fields, Math. Comp. 66 (1997), 1213-1237 A. Derhem, Capitulation dans les extensions quadratiques de corps de nombres cubiques cycliques, Thèse de doctorat, Université Laval, Québec, 1988 D. C. Mayer, Multiplicities of dihedral discriminants, Math. Comp. 58 (1992), 831-847 and S55-S58 |
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E-mail addresses:
Karim.Belabas@math.u-psud.fr aderhem@yahoo.fr danielmayer@algebra.at |
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Yuck ! Here they are -- the Cyclic Cubic Monsters (2002/04/03) |
Dan (02/04/02):
April 2nd, 2002, is the first day since more than 3 months such that I finally became able to really answer Aïssa's very first question about cyclic cubics to me on December 23rd, 2001, one day before Christmas. Today in the early morning, my two fastest machines, having worked simultaneously the whole night, revealed the destiny, fate, will of god, or whatever you prefer. (*) mainframe "HASSE" completed 30000 < f < 40000 at 06:43 (**) mainframe "VORONOI" completed 40000 < f < 50000 at 07:16 My heart stopped beating when I got the results of the SQL-statement "SELECT DISTINCT h FROM table ORDER BY h;": (*) ...,243,... occurs among 30000 < f < 40000 (**) ...,81,...,243,... occur among 40000 < f < 50000 The next question, of course, was "SELECT * FROM table WHERE (h = 81 OR h = 243);" since we want to know which horrible MONSTERS cause such results: (*) a single 243-monster in the doublet f = 36667 = 37*991 with (h-,h+) = (27,243) (**) a single 81-monster in the doublet f = 41977 = 13*3229 with (h-,h+) = (81,189) (189 = 27 * 7) two different monsters in the doublet f = 42127 = 103*409 with (h-,h+) = (243,81) a single 81-monster in the doublet f = 42991 = 13*3307 with (h-,h+) = (27,81) REMARK: h+ refers to the field with x = y (mod 3) h- refers to the field with x = -y (mod 3) in Aïssa's decomposition criteria. Surely this also depends on the choice of the primitive roots g1,g2 and is not an invariant characterization. I thank god that the answers came relatively soon -- around 42000 there is a real accumulation of monsters -- and we didn't have to wait until f ~ 100000 or even higher. |
Dan (02/04/10):
Finally, I give the complete list of all class numbers h divisible by 27 = 3^3 in the range 0 < f < 10^5 of cyclic cubic 2-prime conductors f = q_1*q_2, emphasizing cases with 81 = 3^4 | h in boldface font: f = 4711 = 7*673 with (h-,h+) = (27,27) f = 5383 = 7*769 with (h-,h+) = (27,27) f = 11167 = 13*859 with (h-,h+) = (27,27) f = 12403 = 79*157 with (h-,h+) = (27,27) f = 12439 = 7*1777 with (h-,h+) = (27,27) f = 16177 = 7*2311 with (h-,h+) = (27,27) f = 17593 = 73*241 with (h-,h+) = (189,27) f = 20421 = 9*2269 with (h-,h+) = (189,27) f = 21763 = 7*3109 with (h-,h+) = (27,108) f = 25963 = 7*3709 with (h-,h+) = (27,27) f = 27571 = 79*349 with (h-,h+) = (27,27) f = 28177 = 19*1483 with (h-,h+) = (27,108) f = 32311 = 79*409 with (h-,h+) = (27,108) f = 32689 = 97*337 with (h-,h+) = (27,108) f = 35163 = 9*3907 with (h-,h+) = (189,27) f = 36667 = 37*991 with (h-,h+) = (27,243) f = 37933 = 7*5419 with (h-,h+) = (27,27) f = 38503 = 139*277 with (h-,h+) = (27,108) f = 40573 = 13*3121 with (h-,h+) = (27,189) f = 40873 = 7*5839 with (h-,h+) = (27,27) f = 41977 = 13*3229 with (h-,h+) = (81,189) f = 42127 = 103*409 with (h-,h+) = (243,81) f = 42991 = 13*3307 with (h-,h+) = (27,81) f = 43081 = 67*643 with (h-,h+) = (189,27) f = 44397 = 9*4933 with (h-,h+) = (27,27) f = 49743 = 9*5527 with (h-,h+) = (27,27) f = 49849 = 79*631 with (h-,h+) = (27,108) f = 51847 = 139*373 with (h-,h+) = (27,27) f = 55657 = 7*7951 with (h-,h+) = (27,27) f = 55951 = 7*7993 with (h-,h+) = (27,27) f = 56223 = 9*6247 with (h-,h+) = (27,27) f = 57811 = 13*4447 with (h-,h+) = (27,27) f = 58329 = 9*6481 with (h-,h+) = (27,189) f = 59803 = 79*757 with (h-,h+) = (27,108) f = 59911 = 181*331 with (h-,h+) = (27,27) f = 62257 = 13*4789 with (h-,h+) = (108,351) f = 64971 = 9*7219 with (h-,h+) = (108,27) f = 65383 = 151*433 with (h-,h+) = (27,108) f = 66829 = 7*9547 with (h-,h+) = (108,189) f = 68857 = 37*1861 with (h-,h+) = (81,108) f = 69183 = 9*7687 with (h-,h+) = (27,27) f = 71611 = 19*3769 with (h-,h+) = (27,108) f = 72099 = 9*8011 with (h-,h+) = (189,27) f = 73873 = 31*2383 with (h-,h+) = (27,27) f = 75859 = 7*10837 with (h-,h+) = (27,27) f = 77281 = 109*709 with (h-,h+) = (108,27) f = 78093 = 9*8677 with (h-,h+) = (27,27) f = 81009 = 9*9001 with (h-,h+) = (27,27) f = 84103 = 31*2713 with (h-,h+) = (27,27) f = 89109 = 9*9901 with (h-,h+) = (189,27) f = 89863 = 73*1231 with (h-,h+) = (27,27) f = 94357 = 157*601 with (h-,h+) = (108,27) f = 95913 = 9*10657 with (h-,h+) = (27,27) f = 96709 = 97*977 with (h-,h+) = (27,27) f = 96817 = 7*13831 with (h-,h+) = (27,27) f = 97249 = 79*1231 with (h-,h+) = (81,27) |
Dan (02/04/03):
Now, what about the history of this problem ?
It is an example of a very harmonic cooperation between Aïssa and Dan. None of us would have been able to complete the project alone. But the joint effort made it possible to reach the target. The main stages were the following: |
Aïssa (01/12/23):
Can you help me ?
I would like to know, if there exist cyclic cubic fields of conductor f divisible by two primes and whose 3-class number (i. e., the highest power of 3 dividing the class number h of the field) is 81 (respectively 243). I was told years ago that Professor ENNOLA had tables on these fields with such information and I hope you have access to that tables. |
Dan (01/12/24):
In ENNOLA and TURUNEN's first table (for03.dat.88 from August 11, 1982), which contains the cyclic cubic fields with ordinal numbers 631 <= n <= 1268 and conductors 4003 <= m <= 7999, I found: 2 fields with m = 4711 = 7*673 and both with h = 27 2 fields with m = 5383 = 7*769 and both with h = 27 (These fields occur also in the first additional table of M.-N. GRAS.) In their second table (for03.dat.228 from September 10, 1982), containing the cyclic cubic fields with ordinal numbers 1269 <= n <= 1904 and conductors 8001 <= m <= 11997, I found: 2 fields with m = 11167 = 13*859 and both with h = 27 However, unfortunately I did not find any example of fields whose conductor m has exactly 2 prime factors (whence these fields must necessarily occur as doublets) and where 81 | h, let alone 243 | h. I am quite sure that such cases will appear for bigger values of m. Incidentally, would the discovery of such examples have certain interesting consequences and implications? |
Aïssa (01/12/24):
Now, I study per se the cubic cyclic fields whose conductor is divisible by 2 primes. There exist two such fields for a given conductor and the 3- class number of one of them is 3 (resp 9) if and only if the 3-class number of the other is 3 (resp.9); this is known. I found (the proof is not yet written and is only in my head!) that, if the 3-class number of these two fields is divisible by 27, then it is indeed 27 for one of them (your examples do confirm that case). And what about that of the other, which may be 27, 81, 243,...? Hence my question: is it possible to find such examples? Is there a bound on the value of the 3-class number for such fields? |
Dan (02/03/18):
Today I did succeed in finishing the intricate VORONOI algorithm for totally real cubic fields ;-) Holy moly, I had some damned bugs in the program, unfortunately just for CYCLIC cubics, but now everything with regulators and fundamental systems of units works smooth and fine ! This means we have reached 2 steps in our computational enterprise: 1. TSCHIRNHAUSEN transformed generating polynomials P(X) = X^3 - CX - D with rational integers C,D for the fields K(q1*q2) resp. K(9*q1) 2. Regulators R and fundamental systems (u1,u2) of units for these cyclic cubic fields K Now only the class number h must be determined by means of the analytic class number formula for absolutely abelian number fields (3rd and last step): h = [T * w * sqrt(|disc|)] / [2^s * (2*pi)^t R] where in our case w = number of roots of unity in K = 2 ( only +1 and -1 ) disc = f^2 (square of the conductor) (s,t) = signature = (3,0) (totally real cubic K) and thus h = [T * 2 * f] / [2^3 * R] = T * (f / 4R) Since K is absolutely abelian, the temperament T = lim_{s-->1}(zeta_K(s) / zeta(s)) can be represented in the form T = L(1,chi1) * L(1,chi2) where chi1,chi2 are the non trivial characters on the absolute Galois group Gal(K|Q) of K and the DIRICHLET L-series for complex s with Re(s) >= 1 is given by L(s,chi) = Sum_{n=0}^{infty} [chi(n) / n^s] or as an EULER product L(s,chi) = Prod_{p rational primes} (1 - [chi(p) / p^s])^{-1} Thus the local contribution of a single prime for the temperament T is obviously a product (s = 1): T_p = (1 - [chi1(p) / p])^{-1} * (1 - [chi2(p) / p])^{-1} ================================ ##### PROBLEM: ##### I'm not quite sure yet how to calculate chi1(p) and chi2(p), since Marie Nicole GRAS goes a totally different way. I hope that the character values depend only on the decomposition behavior of the rational prime p. So I suppose what I need is a table of the form decomposition of p in K | chi1(p) | chi2(p) -------------------------|---------|--------- p splits (completely) | ? | ? p remains inert | ? | ? p ramifies | ? | ? Can you help me with this problem? I think it cannot be very difficult! The characters chi1, chi2 should be cubic residue characters or something like that, I conjecture. Maybe you know this stuff by heart. |
Aïssa (02/03/18): You are making real progress toward our computational program. |
Dan (02/03/18):
Yes, revitalizing Voronoi's algorithm in a 4GL for signature (s,t)=(3,0) was a very important step, also for our other possible future projects. (Voronoi for (s,t)=(1,1) is implemented without OOP.) |
Aïssa (02/03/18):
For your question on the computation of the characters,
I think that Gras' paper is the right way to do the computation. |
Dan (02/03/18):
Well, this is true. But now, as I have chosen the way via Voronoi and Euler, I cannot return to M.-N. Gras. Meanwhile I have found my old copy of Shanks' paper "The simplest cubic fields" where he gives the Euler factors for the three decomposition types of primes. So I do not need to deal with the product of 2 L-series. I must only think about how to determine the progressions a + Zf (f...conductor) for split primes. I'm sure that I will soon succeed in the class number computation. |
Aïssa (02/03/19):
Here is how to determine the primes which split
in the cubic fields of conductor f: Soient f le conducteur des 2 corps cubiques cycliques. On a f = q1*q2 où q1 est 9 ou un nombre premier congru à 1 mod 3 et q2 est un premier congru à 1 mod 3. Soit g’1 (resp. g’2) une racine primitive mod q1 (resp. mod q2). Par le théorème des restes chinois il existe des rationnels g1 et g2 tels que {g1 congru à g’1 mod q1 et g1 congru à 1 mod q2} et {g2 congru à 1 mod q1 et g2 congru à g’2 mod q2}. Alors (g1^x)*(g2^y) engendrent le groupe multiplicatif des élément inversibles des entiers mod f lorsque x varie de 1 à 6 si q1= 9 (resp. de 1 à q1-1 si q1 est congru à 1 mod 3) et y varie de 1 à q2-1. Ce groupe est isomorphe au groupe de Galois du corps cyclotomique des racine f-ieme de l’unité sur Q, le corps des nombres rationnels. Dans ce cas l’un des corps cubique est fixé par le sous-groupe des (g1^x)*(g2^y) avec x congru à y mod 3 et l’autre pour ceux pour lesquels x est congru à –y mod 3. Bonne chance et à bientôt |
Dan (02/03/20):
Merci beaucoup, Aïssa!
Now I know that for the given conductor m = q1*q2 the automorphism group of the cyclotomic field Q^m is Gal( Q^m | Q ) ~ U( Z/mZ ) ~ U( Z/q1Z )*U( Z/q2Z ) where, denoting by gi a primitive root mod qi (selected so that g1 = 1 (mod q2), g2 = 1(mod q1) with the aid of the Chinese remainder theorem), the prime residue class groups are U ( Z/qiZ ) ~ { gi^x | 1 <= x <= phi(qi) } for i = 1,2 with phi(qi) = qi-1 for qi = 1 (mod 3) and phi(3^2) = 3*(3-1) = 6, and U ( Z/mZ ) ~ { g1^x1*g2^x2 | 1 <= xi <= phi(qi) for i = 1,2 } Further, you told me that in the field diagram, where L1,L2 denote the 2 fields of conductor m Q^m / \ Q^q1 Q^q2 | / \ | L_q1 L1 L2 L_q2 \ \ / / Q we can distinguish L1 and L2 by means of the subgroups Gal( Q^f | L1 ) ~ { g1^x*g2^y | x = y (mod 3) }, which fixes L1 Gal( Q^f | L2 ) ~ { g1^x*g2^y | x = -y (mod 3) }, which fixes L2 However, I don't see the condition for a prime p to split in L1 or L2. I only know that if p^f = 1 (mod qi), phi(qi) = f*g, then p = P1*...Pg with g prime ideals of degree f in Q^qi for i = 1,2 in particular, p = P1*P2*P3 in L_qi, iff L_qi is contained in the decomposition field Z(Q^qi | Q) of degree g, i. e., iff 3 divides g. Mes amitiés! |
Aïssa (02/03/22):
Si L1 correspond aux éléments (g1^x)*(g2^y) avec x congru à y mod 3, alors un premier rationnel p premier à f se décompose dans L1 si et seulement p = (g1^x)*(g2^y) mod f avec x congru à y mod 3. Pour retrouver votre condition sur la décomposition de p dans Lq1 à savoir que p est un cube mod q1, remarquez que Lq1 correspond aux éléments (g1^x)*(g2^y) avec x congru à 0 mod 3 et donc p se décompose dans Lq1 si et seulement p = (g1^x)*(g2^y) mod f avec x congru à 0 mod 3. Sachant que g2 = 1 mod q1, vous retrouvez que mod q1 la dernière condition devient p est congru à un cube mod q1. Ma condition est plus forte car grâce à y elle décrit la décomposition de p dans le composé de Lq1 et Lq2. A bientôt |
Dan (02/03/22):
1. So, assuming that
Gal( Q^m | L1 ) ~ { g1^x*g2^y | x = y (mod 3) } (i) Gal( Q^m | L2 ) ~ { g1^x*g2^y | x = -y (mod 3) } (ii) the decomposition laws in L1 and L2 (both with conductor m = q1*q2) are: a rational prime p ramifies in L1 and in L2 <==> p divides m splits in L1 <==> p = g1^x*g2^y (mod m) with some x = y (mod 3) remains inert in L1 <==> p incongruent g1^x*g2^y (mod m) for any x = y (mod 3) splits in L2 <==> p = g1^x*g2^y (mod m) with some x = -y (mod 3) remains inert in L2 <==> p incongruent g1^x*g2^y (mod m) for any x = -y (mod 3) With this the Euler product for the class number computation should be done. 2. Concerning the decomposition laws in the fields L_qi, it meanwhile came to my mind that the condition for p to split in L_qi (i = 1 or 2) p^f = 1 (mod qi), phi(qi) = f*g, 3 divides g which I derived from the cyclotomic splitting condition, is obviously equivalent to the Euler criterion for cubic residues p^{phi(qi) / 3} = 1 (mod qi) i. e., as you remarked, p congruent to a cube (mod qi) However, you arrive at this conclusion in an interesting different manner using Gal( Q^m | L_q1 ) ~ { g1^x*g2^y | x = 0 (mod 3) } (iii) Gal( Q^m | L_q2 ) ~ { g1^x*g2^y | y = 0 (mod 3) } (iv) and g2 = 1 mod q1, g1 = 1 mod q2. 3. You told me that you can describe the splitting behavior of p in the compositum L_q1*L_q2 (obviously a field of degree 9). So I assume you mean Gal( Q^m | L_q1*L_q2 ) ~ { g1^x*g2^y | x = 0 (mod 3) and y = 0 (mod 3) } (v) The basic idea with Gal( Q^m | Q ) ~ U( Z/mZ ) ~ { g1^x1*g2^x2 | 1 <= xi <= phi(qi) for i = 1,2 } is contained in M.-N. Gras, but the descriptions (i) to (v) of the subgroups are new for me. Thanks a lot and friendly salutations ! |
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